Proving Cauchy Sequence of |an-bn|

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SUMMARY

If (an) and (bn) are Cauchy sequences, then the sequence (cn) defined as |an - bn| is also a Cauchy sequence. The proof utilizes the triangle inequality, specifically the property that ||X| - |Y|| ≤ |X - Y|. By establishing that |an - am| < ε/2 and |bn - bm| < ε/2 for sufficiently large n and m, it follows that |cn - cm| < ε, confirming that (cn) is Cauchy. This conclusion is critical in understanding the behavior of sequences in metric spaces.

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rbzima
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I'm basically trying to show that if (an) and (bn) are Cauchy sequences, then (cn) = |an - bn| is also a Cauchy sequence.

I know that the triangle inequality is going to be used at one point or another, but I suppose I'm a little confused because:

(an) is Cauchy implies |an - am| < e
(bn) is Cauchy implies |bn - bm| < e

I think at some point my e's are going to be changed to e/2, which is totally legitimate because e is arbitrary anyway.
 
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The key to this is to show that ||X|-|Y||<=|X-Y|

Then ||an-bn|-|am-bm||<=|(an-am)-(bn-bm)|<=|an-am|+|bn-bm|
 

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